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Geometry Difficulty 4.7 AIME Find the answer Canada

In the diagram, $\triangle PQR$ is right-angled at $R$ , $PR=12$ , and $QR=16$ . Also, $M$ is the midpoint of $PQ$ and $N$ is the point on $QR$ so that $MN$ is perpendicular to $PQ$ .The area of $\triangle PNR$ is

Pick one

Solution

Since $\triangle PQR$ is right-angled at $R$ , then by the Pythagorean Theorem, $PQ^2 = PR^2 + QR^2 = 12^2 + 16^2 = 144 + 256 = 400$ Since $PQ>0$ , then $PQ = 20$ . Since $M$ is the midpoint of $PQ$ , then $MQ = \frac{1}{2}PQ = 10$ .

Now $\triangle NMQ$ is similar to $\triangle PRQ$ , since each is right-angled and they share a common angle at $Q$ . Therefore, $\$\dfrac{NQ}{PQ} = \dfrac{MQ}{RQ}$ $and so$ $\dfrac{NQ}{20} = \dfrac{10}{16}$ $which gives$ NQ = $\text{20}$ 10}{16} =
$\dfrac{25}{2}$ $. Thus,$ RN = RQ - NQ = 16 - $\dfrac{25}{2} = \dfrac{32}{2} - \dfrac{25}{2} = \dfrac{7}{2}$ $. Since$ $\triangle$ PNR $is right-angled at$ R $, its area equals$ $\dfrac{1}{2} \cdot PR \cdot$ RN = $\dfrac{1}{2} \cdot 12 \cdot \dfrac{7}{2}$ = 21$.

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